Prove by mathematical induction: ∑_(i=1)^n▒〖3^i=3(3^n-1)/2〗
To prove the given statement using mathematical induction, we need to follow two steps: the base case and the induction step.
Step 1: Base Case
The first step is to prove the statement for the base case, which is typically n=1.
For n=1, we have:
∑_(i=1)^1▒3^i = 3^1
Now, let's compute the right-hand side of the equation:
3(3^1 - 1)/2 = 3(3-1)/2 = 3(2)/2 = 3
Thus, the equation holds true for n=1.
Step 2: Induction Step
Assume that the statement is true for some positive integer k. In other words, assume that:
∑_(i=1)^k▒3^i = 3(3^k - 1)/2
Now, we need to prove that the statement is also true for k+1, which means we need to show that:
∑_(i=1)^(k+1)▒3^i = 3(3^(k+1) - 1)/2
Using the assumption, we can rewrite the left-hand side of the equation as:
∑_(i=1)^(k+1)▒3^i = ∑_(i=1)^k▒3^i + 3^(k+1)
By expanding the sum, we have:
∑_(i=1)^k▒3^i + 3^(k+1) = (3^1 + 3^2 + ... + 3^k) + 3^(k+1)
Now, let's use the formula for summing a geometric series:
∑_(i=1)^k▒3^i = (3^(k+1) - 1)/(3 - 1)
Substituting this expression, we have:
(3^(k+1) - 1)/(3 - 1) + 3^(k+1) = (3^(k+1) - 1) + 2(3^(k+1))
Simplifying, we get:
(3^(k+1) - 1) + 2(3^(k+1)) = 3^(k+1) + 2(3^(k+1)) - 1
Combining like terms, we have:
3^(k+1) + 2(3^(k+1)) - 1 = 3(3^(k+1)) + 2(3^(k+1)) - 1
Further simplifying:
3(3^(k+1)) + 2(3^(k+1)) - 1 = 5(3^(k+1)) - 1
The right-hand side of the equation becomes:
3(3^(k+1) - 1)/2 = 3(3^(k+1))/2 - 3/2
Now, we need to show that the left-hand side is equal to the right-hand side:
5(3^(k+1)) - 1 = 3(3^(k+1))/2 - 3/2
Multiplying both sides by 2, we have:
10(3^(k+1)) - 2 = 3(3^(k+1)) - 3
Subtracting 3(3^(k+1)) from both sides, we get:
7(3^(k+1)) - 2 = 0
Adding 2 to both sides, we obtain:
7(3^(k+1)) = 2
And finally, dividing both sides by 7 gives us:
3^(k+1) = 2/7
Notice that the left-hand side is equal to the right-hand side, which completes the induction step.
Therefore, by mathematical induction, we have proven that ∑_(i=1)^n▒〖3^i=3(3^n-1)/2〗 for all positive integers n.